\(\int \frac {1}{(c-d x^2) \sqrt [3]{c+3 d x^2}} \, dx\) [143]

Optimal result

Integrand size = 23, antiderivative size = 204 \[ \int \frac {1}{\left (c-d x^2\right ) \sqrt [3]{c+3 d x^2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {3} \sqrt {d} x}{\sqrt {c}}\right )}{2\ 2^{2/3} \sqrt {3} c^{5/6} \sqrt {d}}+\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt {d} x}{\sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{c+3 d x^2}\right )}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}}-\frac {\text {arctanh}\left (\frac {\sqrt {c}}{\sqrt {d} x}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}}-\frac {\text {arctanh}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{c+3 d x^2}\right )}{\sqrt {d} x}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}} \]

[Out]

-1/4*arctanh(c^(1/6)*(c^(1/3)-2^(1/3)*(3*d*x^2+c)^(1/3))/x/d^(1/2))*2^(1/3)/c^(5/6)/d^(1/2)-1/4*arctanh(1/x/d^
(1/2)*c^(1/2))*2^(1/3)/c^(5/6)/d^(1/2)-1/12*arctan(x*3^(1/2)*d^(1/2)/c^(1/2))*2^(1/3)/c^(5/6)*3^(1/2)/d^(1/2)+
1/4*arctan(x*3^(1/2)*d^(1/2)/c^(1/6)/(c^(1/3)+2^(1/3)*(3*d*x^2+c)^(1/3)))*3^(1/2)*2^(1/3)/c^(5/6)/d^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {401} \[ \int \frac {1}{\left (c-d x^2\right ) \sqrt [3]{c+3 d x^2}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt {d} x}{\sqrt [6]{c} \left (\sqrt [3]{2} \sqrt [3]{c+3 d x^2}+\sqrt [3]{c}\right )}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}}-\frac {\arctan \left (\frac {\sqrt {3} \sqrt {d} x}{\sqrt {c}}\right )}{2\ 2^{2/3} \sqrt {3} c^{5/6} \sqrt {d}}-\frac {\text {arctanh}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{c+3 d x^2}\right )}{\sqrt {d} x}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}}-\frac {\text {arctanh}\left (\frac {\sqrt {c}}{\sqrt {d} x}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}} \]

[In]

Int[1/((c - d*x^2)*(c + 3*d*x^2)^(1/3)),x]

[Out]

-1/2*ArcTan[(Sqrt[3]*Sqrt[d]*x)/Sqrt[c]]/(2^(2/3)*Sqrt[3]*c^(5/6)*Sqrt[d]) + (Sqrt[3]*ArcTan[(Sqrt[3]*Sqrt[d]*
x)/(c^(1/6)*(c^(1/3) + 2^(1/3)*(c + 3*d*x^2)^(1/3)))])/(2*2^(2/3)*c^(5/6)*Sqrt[d]) - ArcTanh[Sqrt[c]/(Sqrt[d]*
x)]/(2*2^(2/3)*c^(5/6)*Sqrt[d]) - ArcTanh[(c^(1/6)*(c^(1/3) - 2^(1/3)*(c + 3*d*x^2)^(1/3)))/(Sqrt[d]*x)]/(2*2^
(2/3)*c^(5/6)*Sqrt[d])

Rule 401

Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[b/a, 2]}, Simp[q*(ArcTanh
[Sqrt[3]/(q*x)]/(2*2^(2/3)*Sqrt[3]*a^(1/3)*d)), x] + (-Simp[q*(ArcTan[(a^(1/3)*q*x)/(a^(1/3) + 2^(1/3)*(a + b*
x^2)^(1/3))]/(2*2^(2/3)*a^(1/3)*d)), x] + Simp[q*(ArcTan[q*x]/(6*2^(2/3)*a^(1/3)*d)), x] + Simp[q*(ArcTanh[Sqr
t[3]*((a^(1/3) - 2^(1/3)*(a + b*x^2)^(1/3))/(a^(1/3)*q*x))]/(2*2^(2/3)*Sqrt[3]*a^(1/3)*d)), x])] /; FreeQ[{a,
b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[b*c + 3*a*d, 0] && PosQ[b/a]

Rubi steps \begin{align*} \text {integral}& = -\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt {d} x}{\sqrt {c}}\right )}{2\ 2^{2/3} \sqrt {3} c^{5/6} \sqrt {d}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt {d} x}{\sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{c+3 d x^2}\right )}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c}}{\sqrt {d} x}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{c+3 d x^2}\right )}{\sqrt {d} x}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 4.94 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\left (c-d x^2\right ) \sqrt [3]{c+3 d x^2}} \, dx=\frac {3 c x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},-\frac {3 d x^2}{c},\frac {d x^2}{c}\right )}{\left (c-d x^2\right ) \sqrt [3]{c+3 d x^2} \left (3 c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},-\frac {3 d x^2}{c},\frac {d x^2}{c}\right )+2 d x^2 \left (\operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{3},2,\frac {5}{2},-\frac {3 d x^2}{c},\frac {d x^2}{c}\right )-\operatorname {AppellF1}\left (\frac {3}{2},\frac {4}{3},1,\frac {5}{2},-\frac {3 d x^2}{c},\frac {d x^2}{c}\right )\right )\right )} \]

[In]

Integrate[1/((c - d*x^2)*(c + 3*d*x^2)^(1/3)),x]

[Out]

(3*c*x*AppellF1[1/2, 1/3, 1, 3/2, (-3*d*x^2)/c, (d*x^2)/c])/((c - d*x^2)*(c + 3*d*x^2)^(1/3)*(3*c*AppellF1[1/2
, 1/3, 1, 3/2, (-3*d*x^2)/c, (d*x^2)/c] + 2*d*x^2*(AppellF1[3/2, 1/3, 2, 5/2, (-3*d*x^2)/c, (d*x^2)/c] - Appel
lF1[3/2, 4/3, 1, 5/2, (-3*d*x^2)/c, (d*x^2)/c])))

Maple [F]

\[\int \frac {1}{\left (-d \,x^{2}+c \right ) \left (3 d \,x^{2}+c \right )^{\frac {1}{3}}}d x\]

[In]

int(1/(-d*x^2+c)/(3*d*x^2+c)^(1/3),x)

[Out]

int(1/(-d*x^2+c)/(3*d*x^2+c)^(1/3),x)

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (c-d x^2\right ) \sqrt [3]{c+3 d x^2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(-d*x^2+c)/(3*d*x^2+c)^(1/3),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{\left (c-d x^2\right ) \sqrt [3]{c+3 d x^2}} \, dx=- \int \frac {1}{- c \sqrt [3]{c + 3 d x^{2}} + d x^{2} \sqrt [3]{c + 3 d x^{2}}}\, dx \]

[In]

integrate(1/(-d*x**2+c)/(3*d*x**2+c)**(1/3),x)

[Out]

-Integral(1/(-c*(c + 3*d*x**2)**(1/3) + d*x**2*(c + 3*d*x**2)**(1/3)), x)

Maxima [F]

\[ \int \frac {1}{\left (c-d x^2\right ) \sqrt [3]{c+3 d x^2}} \, dx=\int { -\frac {1}{{\left (3 \, d x^{2} + c\right )}^{\frac {1}{3}} {\left (d x^{2} - c\right )}} \,d x } \]

[In]

integrate(1/(-d*x^2+c)/(3*d*x^2+c)^(1/3),x, algorithm="maxima")

[Out]

-integrate(1/((3*d*x^2 + c)^(1/3)*(d*x^2 - c)), x)

Giac [F]

\[ \int \frac {1}{\left (c-d x^2\right ) \sqrt [3]{c+3 d x^2}} \, dx=\int { -\frac {1}{{\left (3 \, d x^{2} + c\right )}^{\frac {1}{3}} {\left (d x^{2} - c\right )}} \,d x } \]

[In]

integrate(1/(-d*x^2+c)/(3*d*x^2+c)^(1/3),x, algorithm="giac")

[Out]

integrate(-1/((3*d*x^2 + c)^(1/3)*(d*x^2 - c)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (c-d x^2\right ) \sqrt [3]{c+3 d x^2}} \, dx=\int \frac {1}{\left (c-d\,x^2\right )\,{\left (3\,d\,x^2+c\right )}^{1/3}} \,d x \]

[In]

int(1/((c - d*x^2)*(c + 3*d*x^2)^(1/3)),x)

[Out]

int(1/((c - d*x^2)*(c + 3*d*x^2)^(1/3)), x)